The answer to this question offers a method for applying initial conditions to a d'Alambert wave equation solution. However, I am still confused by the ambiguity in the standard notation, so I want to go through the solution process making up my own nonstandard notation.
For the purposes of this question, I will stipulate that Lagrange's notation for derivatives (prime symbols) not be used, instead expanding Leibniz's notation ($d$ and $\partial$ symbols), and I will indicate the resulting nonstandard notation by writing it in $\color{blue}{blue}$. The expansion is as follows: Expressions can appear in the "numerator" and/or "denominator" of ordinary and partial derivatives (i.e., if $b = a^4$, then $\frac{dc}{db}$ could be written as $\color{blue}{\frac{dc}{d(a^4)}}$; additionally, a value in parentheses to be plugged in for a variable (or expression) after differentiation also includes the variable, in the form "variable $\rightarrow$ value" (i.e., $\color{blue}{\frac{dc}{d(a^4)}(a^4 \rightarrow 2)}$ more clearly states what might be meant by the more standard but ambiguous $c'(2)$). Now onto the problem.
$$u_{tt} = 4u_{xx},\ u(x,\ 0) = \sin(\pi x),\ u_t (x,\ 0) = 0.$$
I factored the differential operators and used undetermined coefficients to find the general solution $u = h(2t + x) + g(2t - x)$ for arbitrary functions $h$ and $g$. We can infer from this the more general conclusion that $h = h(x,\ t)$ and $g = g(x,\ t)$. Applying the initial conditions yields $$\begin{cases}\sin(\pi x) = h(x) + g(-x) \\ 0 = \frac{\partial}{\partial t}[h(2t + x) + g(2t - x)]_{t = 0}\end{cases}$$
By the chain rule, the second equation becomes $0 = 2\color{blue}{\frac{dh}{d(2t + x)}(t \rightarrow 0)} + 2\color{blue}{\frac{dg}{d(2t - x)}(t \rightarrow 0)}$, easily rewritten as $\color{blue}{\frac{dh}{d(2t + x)}(t \rightarrow 0)} + \color{blue}{\frac{dg}{d(2t - x)}(t \rightarrow 0)} = 0$, but because the functions $h(x,\ t)$ and $g(x,\ t)$ are such that they only contain $t$ in the pattern of $2t + x$ and $2t - x$, respectively, this is the same as $\color{blue}{\frac{dh}{d(2t + x)}((2t + x) \rightarrow x)} + \color{blue}{\frac{dg}{d(2t - x)}((2t - x) \rightarrow -x)} = 0$. This is ostensibly what the linked answer means by the notation $h'(x) + g'(-x) = 0$.
Copying the next step in the linked answer, the substitution $\tilde g(x) = g(-x)$ leads to the system $$\begin{cases}h(x) + \tilde g(x) = \sin(πx) \\ \color{blue}{\frac{dh}{d(2t + x)}((2t + x) \rightarrow x)} - \color{blue}{\frac{d\tilde g}{d(2t - x)}((2t - x) \rightarrow x)} = 0\end{cases}$$
From here the linked answer claims that the second equation in the system antidifferentiates to $h(x) - \tilde g(x) = c$, for some arbitrary constant $c$. This step is unclear to me, and feels like a potential mistake resulting from the oversimplifying Lagrange notation. I am unsure of what I should antidifferentiate with respect to. In my notation, we have $\int \color{blue}{\frac{dh}{d(2t + x)}((2t + x) \rightarrow x)} - \color{blue}{\frac{d\tilde g}{d(2t - x)}((2t - x) \rightarrow x)}\ d? = \int 0\ d?$. My first hypothesis was that I should antidifferentiate both sides with respect to $t$, producing an arbitrary function $f(x)$ on the RHS, because the equation was partially differentiated with respect to $t$ initially. I also considered antidifferentiating with respect to $2t + x$ or $2t - x$, because these are the expressions $h$ and $\tilde g$ are respectively differentiated with respect to currently, but the problem with such an approach is I cannot antidifferentiate different terms in the equation with respect to different expressions. Finally, I could antidifferentiate with respect to $x$, producing an arbitrary function $f(t)$ on the RHS. In no case should the "constant of integration" be a true constant $c$, because although Lagrange's notation makes it look like we're performing single variable antidifferentiation, $h$ and $ \tilde g$ are each still technically functions of both $x$ and $t$. Granted, we are guaranteed that $t$ will not explicitly appear in the functions we're antidifferentiating because plugging in the initial conditions focuses our attention on a particular slice of the overall solution, but this doesn't stop time from existing; the domain of the solution still spans two spacetime dimensions, so taking derivatives with respect to one dimension will make functions of the orthogonal dimension disappear.
How should this equation be antidifferentiated? With respect to what variable or expression is the antiderivative, what is the rational behind this choice, and what is the solution to this antidifferentiated equation? Please do not rely on any understanding being conveyed by Lagrange's notation in the answer.